Simplify and expand the following expression: $ \dfrac{1}{k - 6}+ \dfrac{3}{k + 6}+ \dfrac{3k}{k^2 - 36} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{3k}{k^2 - 36} = \dfrac{3k}{(k - 6)(k + 6)}$ Now we have: $ \dfrac{1}{k - 6}+ \dfrac{3}{k + 6}+ \dfrac{3k}{(k - 6)(k + 6)} $ The least common multiple of the denominators is: $ (k - 6)(k + 6)$ In order to get the first term over $(k - 6)(k + 6)$ , multiply by $\dfrac{k + 6}{k + 6}$ $ \dfrac{1}{k - 6} \times \dfrac{k + 6}{k + 6} = \dfrac{k + 6}{(k - 6)(k + 6)} $ In order to get the second term over $(k - 6)(k + 6)$ , multiply by $\dfrac{k - 6}{k - 6}$ $ \dfrac{3}{k + 6} \times \dfrac{k - 6}{k - 6} = \dfrac{3(k - 6)}{(k - 6)(k + 6)} $ Now we have: $ \dfrac{k + 6}{(k - 6)(k + 6)} + \dfrac{3(k - 6)}{(k - 6)(k + 6)} + \dfrac{3k}{(k - 6)(k + 6)} $ $ = \dfrac{ k + 6 + 3(k - 6) + 3k} {(k - 6)(k + 6)} $ Expand: $ = \dfrac{k + 6 + 3k - 18 + 3k}{k^2 - 36} $ $ = \dfrac{7k - 12}{k^2 - 36}$